Proving that $ab(a+b) = 0_R$ for all elements $a$, $b$ in a ring $R$

Question

Let $R$ be a commutative, Boolean ring with no nonzero nilpotent elements. I want to prove that for all $a$, $b\in R$, $$ab(a+b) = 0_R$$ So what I've done is expressed all powers $n\in \mathbb N$ of $a+b$ as a series expansion as follows: $$(a+b)^n = (a+b)\bigg(\sum_{i=1}^n a^{n-i}b^{i-1}\bigg)$$ But since $R$ is Boolean, then $(a+b)^n = a+b$, thus the series must be equal to $1_R$. There are $n-1$ terms in the series, but all terms other than the $1st$ and final terms are powers of $ab$, which should also be equal to $ab$ by the Boolean property. There are $n-3$ such terms, so I arrived at $$(a+b)^n = (a+b) + ab(2n-4)$$ Which implies that $ab(2n-4) = 0_R$. My claim is that this can only be true for all $n$ provided that $ab = 0_R$. Thus, $ab(a+b) = 0_R (a+b) = 0_R$. Is my thought process here correct? Or is my claim wrong?

Answer 1

Why not simply distribute, use the assumed commutativity, and use the fact that $x^2= x$ and $x + x= 0$ in a Boolean ring: $$ ab(a + b)= aba + abb= aab + abb= a^2b + ab^2= ab + ab= 0 $$